Question 56 – 1.1 Proof by Contradiction - Questions

A security analyst is verifying a property of an encryption key generation algorithm. The analyst needs to prove the following statement:

'For every non-zero rational number $α\alpha$ and every irrational number $β\beta$ , the result of the calculation $5αβ\frac{5\alpha}{\beta}$ is always an irrational number.'

Identify the correct starting assumption for a proof by contradiction of this statement.

There exists a non-zero rational number $α\alpha$ and an irrational number $β\beta$ such that $5αβ\frac{5\alpha}{\beta}$ is rational.
For all non-zero rational numbers $α\alpha$ and all irrational numbers $β\beta$ , the value $5αβ\frac{5\alpha}{\beta}$ is rational.
There exists a non-zero rational number $α\alpha$ and an irrational number $β\beta$ such that $5αβ\frac{5\alpha}{\beta}$ is irrational.
For all non-zero rational numbers $α\alpha$ and all irrational numbers $β\beta$ , the value $5αβ\frac{5\alpha}{\beta}$ is irrational.

1.1 Proof by Contradiction Questions

Practise Edexcel A Level Maths 1.1 Proof by Contradiction with exam-style questions for A Level Maths. 100 questions, matched to the Edexcel A Level Maths (9MA0) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.

A security analyst is verifying a property of an encryption key generation algorithm. The analyst needs to prove the following statement:

'For every non-zero rational number $α\alpha$ and every irrational number $β\beta$ , the result of the calculation $5αβ\frac{5\alpha}{\beta}$ is always an irrational number.'

Identify the correct starting assumption for a proof by contradiction of this statement.

There exists a non-zero rational number $α\alpha$ and an irrational number $β\beta$ such that $5αβ\frac{5\alpha}{\beta}$ is rational.
For all non-zero rational numbers $α\alpha$ and all irrational numbers $β\beta$ , the value $5αβ\frac{5\alpha}{\beta}$ is rational.
There exists a non-zero rational number $α\alpha$ and an irrational number $β\beta$ such that $5αβ\frac{5\alpha}{\beta}$ is irrational.
For all non-zero rational numbers $α\alpha$ and all irrational numbers $β\beta$ , the value $5αβ\frac{5\alpha}{\beta}$ is irrational.